To do this, you can determine how much each angle measures using the cosine law, which is a formula that relates the angles to the sides of any triangle.
[tex]\begin{gathered} a^2=b^2+c^2-2bc\cdot\cos (A) \\ b^2=a^2+c^2-2ac\cdot\cos (B) \\ c^2=a^2+b^2-2ab\cdot\cos (C) \end{gathered}[/tex]
For example, to first find the measure of angle A, you have
[tex]\begin{gathered} a=9,b=8,c=7 \\ a^2=b^2+c^2-2bc\cdot\cos (A) \\ \text{Solve for A and replace the side measurements} \\ a^2+2bc\cdot\cos (A)=b^2+c^2-2bc\cdot\cos (A)+2bc\cdot\cos (A) \\ a^2+2bc\cdot\cos (A)-a^2=b^2+c^2-a^2 \\ 2bc\cdot\cos (A)=b^2+c^2-a^2 \\ \cos (A)=\frac{b^2+c^2-a^2}{2bc} \\ \cos (A)=\frac{8^2+7^2-9^2}{2(7)(8)}=\frac{32}{112} \\ \cos (A)=0.2857 \\ \text{ Apply to both sides of the equation }\cos ^{-1}(x)\text{ which is the inverse function of cos (x)} \\ \cos ^{-1}(\cos (A))=\cos ^{-1}(0.2857)\text{ } \\ A=73.4 \end{gathered}[/tex]
Similarly, you can find the angle B
[tex]\begin{gathered} \cos (B)=\frac{a^2+c^2-b^2}{2ac} \\ \cos (B)=\frac{9^2+7^2-8^2}{2(9)(7)}=\frac{11}{21} \\ \cos (B)=0.5238 \\ \cos ^{-1}(\cos (B))=\cos ^{-1}(0.5238) \\ B=58.4 \end{gathered}[/tex]
To find the angle C you can use the following expression that indicates that the sum of the internal angles of a triangle is 180
[tex]\begin{gathered} A+B+C=180 \\ 73.4+58.4+C=180 \\ C=180-73.4-58.4 \\ C=48.2 \end{gathered}[/tex]
Therefore, the order of the angles from largest to smallest is
